Here's a fun fact from knot theory:

$\quad$ If $\, \Sigma$ is a minimal-genus Seifert surface for a knot $K$, then $i_*:\pi_1(S^3 \setminus \Sigma) \to \pi_1(S^3 \setminus K)$ is injective, where $i: S^3 \setminus \Sigma \to S^3 \setminus K$ is the inclusion.

For a proof, see the "Lemma" in my second answer to this other question. It proves a slightly stronger result: If $\, \Sigma$ is a connected, compact, orientable surface in $S^3$ and $\pi_1(\Sigma^\pm)\to \pi_1(S^3 \setminus \Sigma)$ is injective, then $\pi_1(S^3 \setminus \Sigma) \to \pi_1(S^3 \setminus \partial \Sigma)$ is injective. With the goal of extending the first fact to links with multiple components, I ask:

When does $\pi_1(\Sigma^\pm)$ inject into $\pi_1(S^3 \setminus \Sigma)$? In particular, under what conditions does this hold for a minimal-genus Seifert surface of a link?

Being a minimal-genus Seifert surface is not enough, as is demonstrated by considering the two-component unlink bounding a standard annulus. There are some sufficient conditions:


  1. $\Sigma$ is a minimal Seifert surface for a knot, i.e. the first fact.

  2. $\Sigma$ is a minimal Seifert surface for a two- or three-component link $L$ in which each component has nonzero linking number with some other component. If $\pi_1(\Sigma^+) \to \pi_1(S^3 \setminus \Sigma)$ has nontrivial kernel, the loop theorem gives us a nontrivial curve in $\Sigma^+$ bounding an embedded disk in $S^3 \setminus \Sigma$. Cutting $\Sigma^+$ along the curve and capping with parallel copies of the disk produces an embedded compact orientable surface $\Sigma'$. If $\Sigma'$ is connected, it is a Seifert surface for $L$ with genus less than that of $\Sigma$, a contradiction. If $\Sigma'$ is disconnected and contains a closed component (which will not be $S^2$ by construction), we can remove it to lower genus and get another contradiction. Finally, if $\Sigma'$ is disconnected and contains no closed components, then one of its connected components has only one boundary component and is therefore a Seifert surface for one of the link components, say $L_1$. But since there is some other link component, say $L_2$, whose linking number with $L_1$ is nonzero, $L_2$ must intersect the surface component bounded by $L_1$. This contradicts the fact that $\Sigma'$ is embedded. We conclude that $\pi_1(\Sigma^\pm)\to \pi_1(S^3 \setminus \Sigma)$ is injective.


1 Answer 1


If $\Sigma$ is a minimal Seifert surface of a link $L = \partial \Sigma$ contained in $S^3$, then you can find a taut foliation in the complement of $L$ having $\Sigma$ has leaf (this is a theorem proved by Thurston during the eighties). On the other hand, if $\Sigma$ is a leaf of a taut foliation of a three-manifold $Y$ (in this case the complement of the link), then the inclusion $i : \Sigma \to Y$ induces an injection on the fundamental groups, and we are done.

For more about this techniques, see this survey article by David Gabai: http://www.mathunion.org/ICM/ICM1990.1/Main/icm1990.1.0609.0620.ocr.pdf

If the surface is non minimal this argument does not apply, in fact you can find counterexamples also in the case of knots.

In general, the injectivity condition you are looking for is equivalent to $\textit{incompressibility}$. There is a lot of standard three-manifold theory developed around this condition, see for example this book https://books.google.it/books/about/Algorithmic_and_Computer_Methods_for_Thr.html?id=bjcZAQAAIAAJ&hl=it

  • $\begingroup$ Great. So it looks like the only problem is when the link is split, which is why the claim fails for the unlink. $\endgroup$
    – Kyle
    Jun 4, 2015 at 13:06
  • $\begingroup$ I should note that this is a good answer to the stated question about the injectivity of the map $\pi_1(\Sigma)\to \pi_1(S^3 \setminus \Sigma)$ because $\pi_1(\Sigma) \to \pi_1(S^3 \setminus \partial \Sigma)$ factors through $\pi_1(\Sigma) \to \pi_1(S^3 \setminus \Sigma)$. $\endgroup$
    – Kyle
    Jun 4, 2015 at 13:18
  • $\begingroup$ Oh yes, you where asking for the inclusion $\pi_1( \Sigma) \to \pi_1(S^3 \smallsetminus \Sigma)$. Sure your observation is important! $\endgroup$ Jun 4, 2015 at 23:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .